State the law of conservation of energy in your own words

What kind of energy is in a rock at the edge of a cliff.

soldi70 [24.7K]

Answer:

kinetic energy

Explanation:

8 0

2 years ago

1. A student is biking to school. She travels 0.7 km north, then realizes something has fallen out of her bag.

Snezhnost [94]

Explanation:

(a) Displacement of an object is the shortest path covered by it.

In this problem, a student is biking to school. She travels 0.7 km north, then realizes something has fallen out of her bag. She travels 0.3 km south to retrieve her item. She then travels 0.4 mi north to arrive at school.

0.4 miles = 0.64 km

displacement = 0.7-0.3+0.64 = 1.04 km

(b) Average velocity = total displacement/total time

t = 15 min = 0.25 hour

$v=\dfrac{1.04\ km}{0.25\ h}\\\\v=4.16\ km/h$

Hence, this is the required solution.

8 0

4 years ago

An air conditioner running with R-134a on a cycle executed under the saturationdome between the pressure limits of 0.8 MPa and 0

ohaa [14]

Answer:

The COP of the system is = 4.6

Explanation:

Given data

Higher pressure = 1.8 M pa

Lower pressure = 0.12 M pa

Now we have to find out high & ow temperatures at these pressure limits.

Higher temperature corresponding to pressure 1.8 M pa

$T_{H} = 62.9$ °c = 335.9 K

Lower temperature corresponding to pressure 0.2 M pa

$T_{L} = - 10.1$ °c = 262.9 K

COP of the system is given by

$COP = \frac{T_{L} }{T_{H} -T_{L} }$

$COP = \frac{335.9}{335.9 -262.9}$

COP = 4.6

Therefore the COP of the system is = 4.6

8 0

3 years ago

From the top of a tall building, a gun is fired. The bullet leaves the gun at a speed of 340 m/s, parallel to the ground. As the

Ivahew [28]

Answer:

The launching point is at a distance D = 962.2m and H = 39.2m

Explanation:

It would have been easier with the drawing. This problem is a projectile launching exercise, as they give us data after the window passes and the wall collides, let's calculate with this data the speeds at the point of contact with the window.

X axis

x = Vox t

t = x / vox

t = 7.1 / 340

t = 2.09 10-2 s

In this same time the height of the window fell

Y = Voy t - ½ g t²

Let's calculate the initial vertical speed, this speed is in the window

Voy = (Y + ½ g t²) / t

Voy = [0.6 + ½ 9.8 (2.09 10⁻²)²] /2.09 10⁻² = 0.579 / 0.0209

Voy = 27.7 m / s

We already have the speed at the point of contact with the window. Now let's calculate the distance (D) and height (H) to the launch point, for this we calculate the time it takes to get from the launch point to the window; at this point the vertical speed is Vy2 = 27.7 m / s

Vy = Voy - gt₂

Vy = 0 -g t₂

t₂ = Vy / g

t₂ = 27.7 / 9.8

t₂ = 2.83 s

This is the time it also takes to travel the horizontal and vertical distance

X = Vox t₂

D = 340 2.83

D = 962.2 m

Y = Voy₂– ½ g t₂²

Y = 0 - ½ g t2

H = Y = - ½ 9.8 2.83 2